stopping power of 0 5 1 5mev protons in mylar
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Stopping Power of 0.5-1.5MeV Protons in Mylar Tinyiko S. Maluleke 1 - PowerPoint PPT Presentation

Stopping Power of 0.5-1.5MeV Protons in Mylar Tinyiko S. Maluleke 1 Supervisors: C. Pineda-Vargas 2 M. Msimanga 2 V. Kuzmin 3 1 Stellenbosch University, 2 iThemba Labs, 3 JINR AIM To measure the experimental stopping power for protons


  1. Stopping Power of 0.5-1.5MeV Protons in Mylar Tinyiko S. Maluleke 1 Supervisors: C. Pineda-Vargas 2 M. Msimanga 2 V. Kuzmin 3 1 Stellenbosch University, 2 iThemba Labs, 3 JINR

  2. AIM ● To measure the experimental stopping power for protons in mylar (C 10 H 8 O 4 ). ● To compare the current results with theoretical calculations and previous experimental results. 20090924 South African Summer Student Practice '09 2

  3. INTRODUCTION ● Stopping Power S: S = 1 dE N dx ● It can be expressed as sum of electronic and nuclear stopping power: S = S e  S n 20090924 South African Summer Student Practice '09 3

  4. EXPERIMENTAL SETUP ● Beam : 0.5- 1.5MeV ● Foil Thickness - Mylar: 1.68 μ m ● Primary Target : Pt/C 20090924 South African Summer Student Practice '09 4

  5. RESULTS (RBS Spectrum) ● C 1 – Pt-scattered 1 H + ● C 2 – Pt-scattered 1 H + through the foil. 20090924 South African Summer Student Practice '09 5

  6. RESULTS (Calibration) ● Scattered proton energy: E 1 = KE 0 where kinematic factor K: K = [ ] 2 1  M 2 2   2 − M 1 2 sin 2  M 2 cos  M 1  M 2 θ – Scattering angle M 1 – Projectile mass M 2 – Target nuclei mass 20090924 South African Summer Student Practice '09 6

  7. RESULTS (Stopping Power) ● The energy loss of protons in mylar foil is rather small (less than 20% of E 1 ). ● Stopping power is evaluated at average energy, E avg . 20090924 South African Summer Student Practice '09 7

  8. RESULTS ● Stopping power: S ≈ E  x if  E  0.20 E 1  E = E 1 − E 2 where:  x and is foil thickness, E avg = E 1  E 2 at 2 20090924 South African Summer Student Practice '09 8

  9. CONCLUSION ● Current results deviates from theory and previous results by 5-10%. ● There may be error in thickness of foil. SIMNRA calculations gives mass of foil as 1.82 μ m, which yield good correspondence. ● Therefore mass of foil has to be comfirmed by weighing method. 20090924 South African Summer Student Practice '09 9

  10. THANK YOU! 20090924 South African Summer Student Practice '09 10

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